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gavmur [86]
3 years ago
8

On July 23, 1983, Air Canada Flight 143 required 22,300 kg of jet fuel to fly from Montreal to Edmonton. The density of jet fuel

is 0.803 g/mL, or 1.77 lb/L. The plane had 7682 L of fuel on board in Montreal. The ground crew there multiplied the 7682 L by the factor 1.77(without units) and concluded that they had 13,597 kg of fuel on board and needed an additional 8703 kg by the factor 1.77(again without units) and concluded they needed to add 4916 L of fuel. They added 5000L. On its flight, the plane ran out of fuel and crash-landed near Winnipeg, hundreds of kilometers short of its destination. (there were few injuries and no fatalities.) What mistake did the ground crew make? How much fuel(in L) should they have added before takeoff?
Engineering
1 answer:
Natasha2012 [34]3 years ago
5 0

Answer:

20, 083 L

Explanation:

The mistake was the result of not using units when converting the 7862 l to Kg. They used the density in pounds hence they multiplied by 1.77 Lb/L and obtained 13597 Lb not Kg as they assumed.

To obtain the amount needed to refuel they subtracted this quantity from the 22,300 Kg required for the trip again obtaining the wrong quantity of 8703 Kg and they converted this to liters by dividing the density to get 4916 L and then placed then 5000 L of fuel

The quantity required was

7862 L * 1.77 Lb/L = 13915.74 Lb (pounds not kilos)

then converting this pounds to Kg by multiplying by  0.454 Kg/L one gets

6173 Kg on board

Amount Required

( 22,300 -6173)  :  16127 Kg

16127 Kg/ 0.803 Kg/L =  20083 L

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If the maximum allowable shear stress is 70 MPa, find the shaft diameter needed to transmit 40 kW when the shaft speed is 250 rp
victus00 [196]

Answer:

The diameter is 50mm

Explanation:

The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.

T=(P×60)/(2×pi×N)

T is the Torque

P is the the power to be transmitted by the shaft; 40kW or 40×10³W

pi=3.142

N is the speed of the shaft; 250rpm

T=(40×10³×60)/(2×3.142×250)

T=1527.689Nm

Diameter of a shaft can be obtained from the formula

T=(pi × SS ×d³)/16

Where

SS is the allowable shear stress; 70MPa or 70×10⁶Pa

d is the diameter of the shaft

Making d the subject of the formula

d= cubroot[(T×16)/(pi×SS)]

d=cubroot[(1527.689×16)/(3.142×70×10⁶)]

d=0.04808m or 48.1mm approx 50mm

7 0
3 years ago
Explain combined normal and shear stresses with sketch. Write the general expression for (a) Normal and shear stresses on inclin
Alborosie

Answer:

a) Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

b) principal stress :

б1 = ( бx + бy ) / 2  - \sqrt{}( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

Explanation:

Combined normal stress and shear stress  sketches attached below

The terms in the sketch are :

бx = tensile stress in x direction

бy =  tensile stress in y direction

Txy = y component of shear stress acting on the perpendicular plane to x axis

бn = Normal stress acting on the inclined plane EF

Tn = shear stress acting on the inclined plane EF

A) Normal and shear stresses on inclined plane

Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

B) principal and maximum shear stresses

principal stress :

б1 = ( бx + бy ) / 2  - \sqrt{}( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

6 0
2 years ago
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